$\dfrac{dy}{dt}=-8y$, and $y=1$ when $t=1$. Solve the equation. Choose 1 answer: Choose 1 answer: (Choice A) A $y=e^{-8t}$ (Choice B) B $y=e^{8-8t}$ (Choice C) C $y=e^{8-t}$ (Choice D) D $y=-8e^{-t}$
The general solution of equations of the form $\dfrac{dy}{dt}=ky$ is $y=C\cdot e^{kt}$ for some constant $C$. This can be found using separation of variables. In our case, $k=-8$, so $y=C\cdot e^{-8t}$. Let's use the fact that $y=1$ when $t=1$ to find $C$ : $\begin{aligned} y&=C\cdot e^{-8t} \\\\ 1&=C\cdot e^{-8\cdot 1} \gray{\text{Plug }t=1\text{ and }y=1} \\\\ e^8&=C \end{aligned}$ In conclusion, $y=e^{8-8t}$.